<span>89 is the correct answer.
</span>
Answer:
a) ![\mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}](https://tex.z-dn.net/?f=%5Cmathrm%7BE%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7B%5Cmathrm%7BH%7D%7D%5E%7B5%7D%20%5Cfrac%7B200%7D%7B101-i%7D)
b) ![\mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bk%3D1%7D%5E%7B5%7D%20%5Cfrac%7B%28200%29%5E%7B2%7D%7D%7B%28101-i%29%5E%7B2%7D%7D)
Step-by-step explanation:
Given:
The lifetimes of the individual items are independent exponential random variables.
Mean = 200 hours.
Assume, Ti be the time between ( 
)st and the 
failures. Then, the 
are independent with 
being exponential with rate 
Therefore,
a) ![E[T]=\sum_{i=1}^{5} E\left[\tau_{i}\right]](https://tex.z-dn.net/?f=E%5BT%5D%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%20E%5Cleft%5B%5Ctau_%7Bi%7D%5Cright%5D)
![\therefore \mathrm{E}[\mathrm{T}]=\sum_{\mathrm{H}}^{5} \frac{200}{101-i}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cmathrm%7BE%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7B%5Cmathrm%7BH%7D%7D%5E%7B5%7D%20%5Cfrac%7B200%7D%7B101-i%7D)
The variance is given by, ![\mathrm{Var}[\mathrm{T}]=\sum_{i=1}^{5} \mathrm{Var}[T]](https://tex.z-dn.net/?f=%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bi%3D1%7D%5E%7B5%7D%20%5Cmathrm%7BVar%7D%5BT%5D)
![\therefore \mathrm{Var}[\mathrm{T}]=\sum_{k=1}^{5} \frac{(200)^{2}}{(101-i)^{2}}](https://tex.z-dn.net/?f=%5Ctherefore%20%5Cmathrm%7BVar%7D%5B%5Cmathrm%7BT%7D%5D%3D%5Csum_%7Bk%3D1%7D%5E%7B5%7D%20%5Cfrac%7B%28200%29%5E%7B2%7D%7D%7B%28101-i%29%5E%7B2%7D%7D)
Answer:
The product of a non-zero rational number and an irrational number is irrational.
Hey there! :D
Since all the terms are being added together, and there is a common denominator with the fractions, we can just add them.
x+x= 2x 3/4+2/4= 5/4
2x+5/4
2x+1.25<== simplest form
I hope this helps!
~kaikers
